Difference between revisions of "MA598R 7.8" - Rhea

Latest revision as of 05:54, 11 June 2013

Problem #7.8, MA598R, Summer 2009, Weigel

Slaughter a horde of pirates to get back to The_Ninja's_Solutions

Prove that $ *:L^{p}(\mathbb{R}^n)\times L^{q}(\mathbb{R}^n)\rightarrow C(\mathbb{R}^n) $ is well defined, if $ 1/p+1/q=1, 1\le p\le\infty $

$ f*g = g*f $ so WOLOG take $ p<\infty $

Let $ \epsilon>0 $

$ \exists h \in C_{0}(\mathbb{R}^n) $ s.t. $ \left|\left|f-h\right|\right|_{p}<\epsilon $

$ h(x) $ has compact support, so it is uniformly continuous and $ \exists r $ s.t. $ h(x) = 0 \forall x, |x|>r $. Uniform continuity implies $ \exists\delta>0 $ s.t. $ \left|x-x'\right|<\delta\Rightarrow\left|h(x)-h(x')\right|<\frac{\epsilon}{\mu(|x|\le r)^{1/p}} $.

Using this $ \delta $, let $ \left|x-x'\right|<\delta $.

$ \left|(f*g)(x)-(f*g)(x')\right| = $

$ = \left|\int_{\mathbb{R}^n}f(x-y)g(y)dy-\int_{\mathbb{R}^n}f(x'-y)g(y)dy\right| $

$ = \left|\int_{\mathbb{R}^n}(f(x-y)-f(x'-y))g(y)dy\right| $

$ \le \left|\left|(f(x-y)-f(x'-y))\right|\right|_{p} \left|\left|g\right|\right|_{q} $ by Holder's Inequality

$ \le (\left|\left|(f(x-y)-h(x-y))\right|\right|_{p}+\left|\left|(h(x-y)-h(x'-y))\right|\right|_{p}+\left|\left|(h(x'-y)-f(x'-y))\right|\right|_{p}) \left|\left|g\right|\right|_{q} $ by Minkowski's Inequality

$ \left|\left|(h(x-y)-h(x'-y))\right|\right|_{p} = \left(\int_{|x|\le r}\left|(h(x-y)-h(x'-y))\right|^{p}dx \right)^\frac{1}{p} < \left(\int_{|x|\le r}\frac{\epsilon^{p}}{\mu(|x|\le r)}dx \right)^\frac{1}{p} =\epsilon $

$ < (\epsilon + \epsilon + \epsilon)\left|\left|g\right|\right|_{q} $

$ \epsilon $ is arbitrary so $ f*g $ is continuous.

~Ben Bartle

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